Equal or not?

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Here we see a dialogue where students discuss combining limits with arithmetic.

Check out this dialogue between two calculus students (based on a true story):

Devyn: Riley, I’ve been thinking about limits.
Riley: So awesome!
Devyn: Think about $\lim _{x\to a} \left (f(x) + g(x)\right ).$ This is the number that $f(x) + g(x)$ gets nearer and nearer to, as $x$ gets nearer and nearer to $a$ .
Riley: You know it!
Devyn: So I think it is the same as $\lim _{x\to a} f(x) + \lim _{x\to a}g(x).$
Riley: Yeah, that does make sense, since when you add two numbers, say $(\text {a number near $6$}) + (\text {a number near $7$})$ you get $(\text {a number near $13$})$
Riley: Right! And I think the same reasoning will work for multiplication! So we should be able to say $\lim _{x\to a}\left (f(x) \cdot g(x)\right ) = \left (\lim _{x\to a} f(x) \right )\cdot \left (\lim _{x\to a} g(x)\right ).$
Devyn: Yes, I think that’s right! But what about division? Can we use similar reasoning to conclude $\lim _{x\to a} \frac {f(x)}{g(x)} = \frac {\lim _{x\to a} f(x)}{\lim _{x\to a} g(x)}.$

Give an argument (similar to the one above) supporting the idea that $\lim _{x\to a}\left (f(x) \cdot g(x)\right ) = \left (\lim _{x\to a} f(x) \right )\cdot \left (\lim _{x\to a} g(x)\right ).$

For the next problems, suppose $L$ is a number near $1$ and that $M$ is a number near $0$ .

Using the context above, $\frac {\text {large}}{\text {small}} = ?$

“large” “small” impossible to say

Using the context above, $\frac {\text {small}}{\text {small}} = ?$

“large” “small” impossible to say